(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Biodiversity Heritage Library | Children's Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "The Flow Of Gases In Furnaces"

DESIGN OF OPEN-HEARTH FURNACES

257

TABLE 13
TABLE GIVING THE VALUES OF A FOR ENGLISH UNITS

r      Tj'rtyv*-
	Values of B
				
hi, .beet
	3 Feet
	6 Feet
	9 Feet
	12 Feet
	16 Feet

1.0
	2.275
	2.36
	2.40
	2.42
	2.43

1.5
	2.215
	2.32
	2.38
	2.40
	2.41

2.0
	2.15
	2.29
	2.35
	2.38
	2.39

2.5
	2.08
	2.245
	2.32
	2.36
	2.38

3.0
	2.01
	2.20
	2.30
	2.35
	2.375

For the case in hand,

Q, = Q(l+orf)=  12 m3 77X7.606 = 97 m3 14

451 cu ft   X 7.606 = 3430 cu ft
5 = 5 m 00 (16.4 ft) or 4 m 650 (15.25 ft)
Z=1800C. (3275 F.)

A = 3.53 and 3.48 (approximately for metric units)
= 2.37 and 2.34 (approximately for English units)

The formula may now be written with numerical values, as
follows:

For 5=5m 00 (16.4ft),

Or, for English units,
For 5=15.25 ft (4m 65),

, = 2.34^-

34302

15.252X1800

= 7.11 ft (2m 169).

This will be the distance from the surface of the bath to the
center of gravity of the roof segment; it will give an approximate
height of the skewbacks above the door sills of 5 ft 4 in (1630 mm)
for the wider chamber and 5 ft 7 in (1700 mm) for the narrower
chamber.
With a chamber area of 65 m2 00 (700 sq ft) the chamber
volume will be approximately 144 m3 00 (5087 cu ft) for the